Difference between revisions of "2002 AMC 10A Problems/Problem 20"
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<math>\text{(A)}\ 5/4 \qquad \text{(B)}\ 4/3 \qquad \text{(C)}\ 3/2 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 2</math> | <math>\text{(A)}\ 5/4 \qquad \text{(B)}\ 4/3 \qquad \text{(C)}\ 3/2 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 2</math> | ||
+ | |||
+ | [asy] | ||
+ | pair A,B,C,D,EE,F,G,H,J; | ||
+ | A = (0,0); | ||
+ | B = (0.2,0); | ||
+ | C = 2*B; | ||
+ | D = 3*B; | ||
+ | EE = 4*B; | ||
+ | F = 5*B; | ||
+ | G = (-0.2,0.8); | ||
+ | H = intersectionpoint(G--D,C -- (C + G)); | ||
+ | J = intersectionpoint(G--F,EE--(EE+G)); | ||
+ | draw(G--F--A--G--B); | ||
+ | draw(H--C--G--D); | ||
+ | draw(J--EE--G); | ||
+ | label("<math>A</math>",A,SW); | ||
+ | label("<math>B</math>",B,S); | ||
+ | label("<math>C</math>",C,S); | ||
+ | label("<math>D</math>",D,S); | ||
+ | label("<math>E</math>",EE,S); | ||
+ | label("<math>F</math>",F,SE); | ||
+ | label("<math>J</math>",J,NE); | ||
+ | label("<math>G</math>",G,N); | ||
+ | label(scale(0.9)*"<math>H</math>",H,NE,UnFill(0.1mm)); | ||
+ | [/asy] | ||
==Solution== | ==Solution== |
Revision as of 20:30, 5 November 2015
Problem
Points and lie, in that order, on , dividing it into five segments, each of length 1. Point is not on line . Point lies on , and point lies on . The line segments and are parallel. Find .
[asy] pair A,B,C,D,EE,F,G,H,J; A = (0,0); B = (0.2,0); C = 2*B; D = 3*B; EE = 4*B; F = 5*B; G = (-0.2,0.8); H = intersectionpoint(G--D,C -- (C + G)); J = intersectionpoint(G--F,EE--(EE+G)); draw(G--F--A--G--B); draw(H--C--G--D); draw(J--EE--G); label("",A,SW); label("",B,S); label("",C,S); label("",D,S); label("",EE,S); label("",F,SE); label("",J,NE); label("",G,N); label(scale(0.9)*"",H,NE,UnFill(0.1mm)); [/asy]
Solution
Solution #1: Since and are parallel, triangles and are similar. Hence, .
Since and are parallel, triangles and are similar. Hence, . Therefore, . The answer is (D).
Solution #2: As is parallel to , angles FJE and FGA are congruent. Also, angle F is clearly congruent to itself. From AA similarity, ; hence . Similarly, . Thus, .
See Also
2002 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.